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1248. Count Number of Nice Subarrays
Description
Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it.
Return the number of nice sub-arrays.
Example 1:
Input: nums = [1,1,2,1,1], k = 3 Output: 2 Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
Example 2:
Input: nums = [2,4,6], k = 1 Output: 0 Explanation: There is no odd numbers in the array.
Example 3:
Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2 Output: 16
Constraints:
1 <= nums.length <= 500001 <= nums[i] <= 10^51 <= k <= nums.length
Solutions
Solution 1: Prefix Sum + Array or Hash Table
The problem asks for the number of subarrays that contain exactly $k$ odd numbers. We can calculate the number of odd numbers $t$ in each prefix array and record it in an array or hash table $cnt$. For each prefix array, we only need to find the number of prefix arrays with $t-k$ odd numbers, which is the number of subarrays ending with the current prefix array.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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class Solution { public int numberOfSubarrays(int[] nums, int k) { int n = nums.length; int[] cnt = new int[n + 1]; cnt[0] = 1; int ans = 0, t = 0; for (int v : nums) { t += v & 1; if (t - k >= 0) { ans += cnt[t - k]; } cnt[t]++; } return ans; } } -
class Solution { public: int numberOfSubarrays(vector<int>& nums, int k) { int n = nums.size(); vector<int> cnt(n + 1); cnt[0] = 1; int ans = 0, t = 0; for (int& v : nums) { t += v & 1; if (t - k >= 0) { ans += cnt[t - k]; } cnt[t]++; } return ans; } }; -
class Solution: def numberOfSubarrays(self, nums: List[int], k: int) -> int: cnt = Counter({0: 1}) ans = t = 0 for v in nums: t += v & 1 ans += cnt[t - k] cnt[t] += 1 return ans -
func numberOfSubarrays(nums []int, k int) (ans int) { n := len(nums) cnt := make([]int, n+1) cnt[0] = 1 t := 0 for _, v := range nums { t += v & 1 if t >= k { ans += cnt[t-k] } cnt[t]++ } return } -
function numberOfSubarrays(nums: number[], k: number): number { const n = nums.length; const cnt = new Array(n + 1).fill(0); cnt[0] = 1; let ans = 0; let t = 0; for (const v of nums) { t += v & 1; if (t - k >= 0) { ans += cnt[t - k]; } cnt[t] += 1; } return ans; }